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I have the acceleration as a function of distance, $a(t)$ $$a(t) = f(S)$$ $$\int v.dv = \int f(S).dS$$

And so I have velocity as a function of time if I want it.

What I need is to find $S(t)$.
I tried to do it using the equation $S=ut+\frac{1}{2}at^2$, but then I realized that this may not be a case of constant acceleration.

PS: Converting a function for "velocity vs. position", $v(x)$, to "position vs. time", $p(t)$ seems to be similar, but I don't understand it/it's not applicable I suppose.

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  • $\begingroup$ How do you solve harmonic motion equation $x''(t)=f(x)=-kx$? $\endgroup$ – Shuchang Nov 3 '13 at 6:21
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I assume this is 1-D motion. In general you could use $\dfrac{d^2s}{dt^2} = f(s)$, and you can solve for $s(t)$.

Or you could break it into two steps, by solving for $v$ first using $a = v \dfrac{dv}{ds} = f(s)$ which is easily separable, then solving for $s$ using $v = \dfrac{ds}{dt} = f(s)$.

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